Question

A Ferris wheel is 10 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 4 minutes. How many minutes of the ride are spent higher than 11 meters above the ground?

Answer 1

make a sketch of the positions at t = 0, 1, 2, 3, and 4
and you will see that you need something like
height = 5sin(π/2)t + 8

to meet your condition of 6:00 o'clock position being 3 m
we do a 1 second shift to the right,
height = 5sin [(π/2)(t-1)] + 8

check: when t = 0 , height = 5sin( -π/2)+8 = 3
when t = 1, height = 8
when t = 2, height = 5sin π/2 + 8 = 13
..
when t=4, height = 5sin π + 8 = 3

equation is correct,
so we want
5sin [(π/2)(t-1)] + 8≥11
(lets use the =)
5sin [(π/2)(t-1)] + 8 = 11
5sin [(π/2)(t-1)] = 3
sin [(π/2)(t-1)] = .6
I know sin (.6435) = appr .6

π/2(t-1) = .6435 or π/2(t-1) = π-.6435
t-1 =.4096655.. or t-1 = 1.59033..
t = appr 1.41 or t = appr 2.59

so time spent higher than 11 m is
2.59-1.41 seconds or
appr 1.18 seconds